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Chapter 2

Fundamentals of the Mechanical
Behavior of Materials
that as the specimen’s gage length is increased,
the contribution of localized elongation (that is,
necking) will decrease, but the total elongation
will not approach zero.

Questions
2.1 Can you calculate the percent elongation of
materials based only on the information given
in Fig. 2.6? Explain.

2.3 Explain why the difference between engineering
strain and true strain becomes larger as strain
increases. Is this phenomenon true for both
tensile and compressive strains? Explain.

Recall that the percent elongation is defined by
Eq. (2.6) on p. 33 and depends on the original
gage length (lo) of the specimen. From Fig. 2.6

The difference between the engineering and true
strains becomes larger because of the way the

area of the specimen increases as the specimen
height decreases (because of volume constancy)
as the load is increased. Since true stress is
defined as ratio of the load to the instantaneous
cross-sectional area of the specimen, the true
stress in compression will be lower than the
1


engineering stress for a given load, assuming
that friction between the platens and the
specimen is negligible.
2.5 Which of the two tests, tension or compression,
requires a higher capacity testing machine than
the other? Explain.
The compression test requires a higher capacity
machine because the cross-sectional area of the
specimen increases during the test, which is the
opposite of a tension test. The increase in area
requires a load higher than that for the tension
test to achieve the same stress level.
Furthermore, note that compression-test
specimens generally have a larger original crosssectional area than those for tension tests, thus
requiring higher forces.
2.6 Explain how the modulus of resilience of a
material changes, if at all, as it is strained: (1)
for an elastic, perfectly plastic material, and (2)
for an elastic, linearly strain-hardening
material.
2.7 If you pull and break a tension-test specimen

of the specimen between the gage lengths, we
obtain the average specific work throughout
the specimen. However, the area under the true
stress-true strain curve represents the specific
work done at the necked (and fractured) region
in the specimen where the strain is a maximum.
Thus, the answers will be different. However,
up to the onset of necking (instability), the
specific work calculated will be the same. This
is because the strain is uniform throughout the
specimen until necking begins.
2.10 The note at the bottom of Table 2.5 states that as
temperature increases, C decreases and m
increases. Explain why.
The value of C in Table 2.5 on p. 43 decreases
with temperature because it is a measure of the
strength of the material. The value of m
increases with temperature because the
material becomes more strain-rate sensitive,
due to the fact that the higher the strain rate, the
less time the material has to recover and
recrystallize, hence its strength increases.
2.11 You are given the K and n values of two different
materials. Is this information sufficient to
determine which material is tougher? If not,
what additional information do you need, and
why?
Although the K and n values may give a good
estimate of toughness, the true fracture stress
and the true strain at fracture are required for

greater length before it fails; this behavior is an
indication that necking is delayed with
increasing m. When necking is about to begin,
the necking region’s strength with respect to
the rest of the specimen increases, due to strain
hardening. However, the strain rate in the
necking region is also higher than in the rest of
the specimen, because the material is
elongating faster there. Since the material in the
necked region becomes stronger as it is
strained at a higher rate, the region exhibits a
greater resistance to necking. The increase in
resistance to necking thus depends on the
magnitude of m. As the tension test progresses,
necking becomes more diffuse, and the
specimen becomes longer before fracture;
hence, total elongation increases with
increasing values of m (Fig. 2.13 on p. 45). As
expected, the elongation after necking
(postuniform elongation) also increases with
3

increasing m. It has been observed that the
value of m decreases with metals of increasing
strength.
2.15 Explain why materials with high m values (such
as hot glass and silly putty) when stretched
slowly, undergo large elongations before
failure. Consider events taking place in the
necked region of the specimen.

versus foil; (d) the size of the specimen with
respect to that of the indenter; and (e) the
surface finish of the part being tested.


2.19 Which hardness tests and scales would you use
for very thin strips of material, such as
aluminum foil? Why?

2.21 In a Brinell hardness test, the resulting
impression is found to be an ellipse. Give
possible explanations for this phenomenon.

Because aluminum foil is very thin, the
indentations on the surface must be very small
so as not to affect test results. Suitable tests
would be a microhardness test such as Knoop
or Vickers under very light loads (see Fig. 2.22
on p. 52). The accuracy of the test can be
validated by observing any changes in the
surface appearance opposite to the indented
side.

There are several possible reasons for this
phenomenon, but the two most likely are
anisotropy in the material and the presence of
surface residual stresses in the material.

2.20 List and explain the factors that you would
consider in selecting an appropriate hardness


4

2.21 Referring to Fig. 2.22 on p. 52, note that the
material for indenters are either steel, tungsten
carbide, or diamond. Why isn’t diamond used
for all of the tests?
While diamond is the hardest material known,
it would not, for example, be practical to make
and use a 10-mm diamond indenter because the
costs would be prohibitive. Consequently, a
hard material such as those listed are sufficient
for most hardness tests.
2.22 What effect, if any, does friction have in a
hardness test? Explain.
The effect of friction has been found to be
minimal. In a hardness test, most of the
indentation
occurs
through
plastic
deformation, and there is very little sliding at
the indenter-workpiece interface; see Fig. 2.25
on p. 55.
2.23 Describe the difference between creep and
stress-relaxation phenomena, giving two
examples for each as they relate to engineering
applications.
Creep is the permanent deformation of a part
that is under a load over a period of time,

propagate as they would when under tensile
stresses. Consequently, the specimens would
basically behave as if they were not notched.
2.25 If you remove layer ad from the part shown in
Fig. 2.30d, such as by machining or grinding,
which way will the specimen curve? (Hint:
Assume that the part in diagram (d) can be
modeled as consisting of four horizontal
springs held at the ends. Thus, from the top
down, we have compression, tension,
compression, and tension springs.)
Since the internal forces will have to achieve a
state of static equilibrium, the new part has to
bow downward (i.e., it will hold water). Such
residual-stress patterns can be modeled with a
set of horizontal tension and compression
springs. Note that the top layer of the material
ad in Fig. 2.30d on p. 60, which is under
compression, has the tendency to bend the bar
upward. When this stress is relieved (such as by
removing a layer), the bar will compensate for
it by bending downward.
2.26 Is it possible to completely remove residual
stresses in a piece of material by the technique
described in Fig. 2.32 if the material is elastic,
linearly strain hardening? Explain.
By following the sequence of events depicted in
Fig. 2.32 on p. 61, it can be seen that it is not
possible to completely remove the residual
stresses. Note that for an elastic, linearly strain

(c) Shoe sole: The sole should be compliant
for comfort, with a high resilience. It
should be tough so that it absorbs shock
and should have high friction and wear
resistance.
(d) Fish hook: A fish hook needs to have high
strength so that it doesn’t deform
permanently under load, and thus
maintain its shape. It should be stiff (for
better control during its use) and should
be resistant the environment it is used in
(such as salt water).
(e) Automotive piston: This product must
have high strength at elevated
temperatures, high physical and thermal
shock resistance, and low mass.
(f) Boat propeller: The material must be stiff
(to maintain its shape) and resistant to
corrosion, and also have abrasion
resistance
because
the
propeller


encounters sand and other abrasive
particles when operated close to shore.
(g) Gas turbine blade: A gas turbine blade
operates at high temperatures (depending
on its location in the turbine); thus it

loss and that the temperature distribution is
uniform throughout. If the specific heat of the
material
decreases
with
increasing
temperature, will the work of deformation
calculated using the specific heat at room
temperature be higher or lower than the actual
work done? Explain.
If we calculate the heat using a constant specific
heat value in Eq. (2.65) on p. 73, the work will
be higher than it actually is. This is because, by
definition, as the specific heat decreases, less
6

work is required to raise the workpiece
temperature by one degree. Consequently, the
calculated work will be higher than the actual
work done.
2.31 Explain whether or not the volume of a metal
specimen changes when the specimen is
subjected to a state of (a) uniaxial compressive
stress and (b) uniaxial tensile stress, all in the
elastic range.
For case (a), the quantity in parentheses in Eq.
(2.47) on p. 69 will be negative, because of the
compressive stress. Since the rest of the terms
are positive, the product of these terms is
negative and, hence, there will be a decrease in



be subjected to tensile radial stresses.
Thus, a state of triaxial (though not exactly
hydrostatic) tension will exist within the
thin disk.
2.33 Referring to Fig. 2.19, make sketches of the
state of stress for an element in the reduced
section of the tube when it is subjected to (1)
torsion only, (2) torsion while the tube is
internally pressurized, and (3) torsion while
the tube is externally pressurized. Assume that
the tube is closed end.
These states of stress can be represented simply
by referring to the contents of this chapter as
well as the relevant materials covered in texts
on mechanics of solids.
2.34 A penny-shaped piece of soft metal is brazed to
the ends of two flat, round steel rods of the
same diameter as the piece. The assembly is
then subjected to uniaxial tension. What is the
state of stress to which the soft metal is
subjected? Explain.
The penny-shaped soft metal piece will tend to
contract radially due to the Poisson’s ratio;
however, the solid rods to which it attached will
prevent this from happening. Consequently, the
state of stress will tend to approach that of
hydrostatic tension.
2.35 A circular disk of soft metal is being compressed

those on the yield locus are in a plastic state.
Points outside the yield locus are not
admissible. Therefore, an increase in σ1 while
the other stresses remain unchanged would
require an increase in yield stress. This can also
be deduced by inspecting either Eq. (2.36) or
Eq. (2.37) on p. 64.
2.37 What is the dilatation of a material with a
Poisson’s ratio of 0.5? Is it possible for a
material to have a Poisson’s ratio of 0.7? Give a
rationale for your answer.
It can be seen from Eq. (2.47) on p. 69 that the
dilatation of a material with ν = 0.5 is always
zero, regardless of the stress state. To examine
the case of ν = 0.7, consider the situation where
the stress state is hydrostatic tension. Equation
(2.47) would then predict contraction under a
tensile stress, a situation that cannot occur.
2.38 Can a material have a negative Poisson’s ratio?
Explain.
Solid material do not have a negative Poisson’s
ratio, with the exception of some composite
materials (see Chapter 10), where there can be
a negative Poisson’s ratio in a given direction.
2.39 As clearly as possible, define plane stress and
plane strain.
Plane stress is the situation where the stresses
in one of the direction on an element are zero;
plane strain is the situation where the strains in
one of the direction are zero.

student is encouraged to develop as many as
possible. Two possible answers are: (1) there is
a tradeoff between mathematical complexity
and accuracy in modeling material behavior
and (2) some materials may be better suited for
certain constitutive laws than others.

0

200 400 600 800 1000 1200

Elastic modulus (GPa)

Typical comments regarding such a chart are:
(a) There is a smaller range for metals than
for non-metals;
(b) Thermoplastics, thermosets and rubbers
are orders of magnitude lower than
metals and other non-metals;

2.42 Plot the data in Table 2.1 on a bar chart,
showing the range of values, and comment on
the results.

(c) Diamond and ceramics can be superior to
others, but ceramics have a large range of
values.

By the student. An example of a bar chart for the
elastic modulus is shown below.

is because the stress-strain curve is not linearly
proportional after the proportional limit, which
can be as high as one-half the yield strength in
some metals. Therefore, a transition from
elastic to plastic behavior in a stress-strain
curve is difficult to discern. The use of a 0.2%
offset is a convenient way of consistently
interpreting a yield point from stress-strain
curves.
2.45 Referring to Question 2.44, would the offset
method be necessary for a highlystrainedhardened material? Explain.
The 0.2% offset is still advisable whenever it
can be used, because it is a standardized
approach for determining yield stress, and thus
one should not arbitrarily abandon standards.
However, if the material is highly cold worked,
there will be a more noticeable ‘kink’ in the
stress-strain curve, and thus the yield stress is
far more easily discernable than for the same
material in the annealed condition.

9


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gone.
Assuming volume constancy, we may write

Problems

radial strain is
in the
three principal directions is zero, indicating
volume constancy in plastic deformation.

2.48 A material has the following properties: UTS =
50,000 psi and n = 0.25 Calculate its strength
coefficient K.
Therefore the true strains are additive. Using the
same approach for engineering strain as defined
Let us first note that the true UTS of this material
by Eq. (2.1), we obtain e1 = 0.1667, e2 = 0.1429,
is given by UTStrue = Knn (because at necking
and e3 = 0.5. The sum of these strains is e1+e2+e3
We can then determine the value of this
= 0.8096. The engineering strain from step 1 to
stress
from
the UTS by following a procedure
3 is
similar to Example 2.1. Since n = 0.25, we can
write
Note that this is not equal to the sum of the
engineering strains for the individual steps.
2.47 A paper clip is made of wire 1.20-mm in
diameter. If the original material from which the
wire is made is a rod 15-mm in diameter,
calculate the longitudinal and diametrical
engineering and true strains that the wire has
under10

found to be about 60,000 psi, while the true (a) Aneck = Aoe−n):
strain is about 0.2. We also know that the ratio
AA = (7)e−0.3 = 5.18 mm2
of the original to necked areas of the specimen is
given by
AB = (2.5)e−0.3 = 1.85 mm2
AC = (3)e−0.3 = 2.22 mm2
or

AD = (2)e−0.3 = 1.48 mm2

Aneck

−0.20

=e

Hence the total load that the cable can support is

= 0.819

P

A

= (314)(5.18) + (418)(1.85)
+(209)(2.22) + (530)(1.48)

o Thus,



.

If the n values of the four strands were different, the
procedure would consist of plotting the loadelongation curves of the four strands on the same
chart, then obtaining graphically the maximum load.
Alternately, a computer program can be written to
determine the maximum load.
Thus
2.52 Using only Fig. 2.6, calculate the maximum load in
tension testing of a 304 stainless-steel
2.51 A cable is made of four parallel strands of difround specimen with an original diameter of 0.5
ferent
materials,
all
behaving
according
to the in.
equation, where n = 0.3 The materials, strength coefficients, and cross sections are We observe from Fig. 2.6 on p. 37 that necking as follows: begins
at a true strain of about 0.1, and that the true UTS is about 110,000 psi. The origiMaterial A: K = 450 MPa, Ao = 7
mm2; nal cross-sectional area is Ao = π(0.25 in)2 =
2

Material B: K = 600 MPa, Ao = 2.5 mm2;

0.196 in . Since n = 0.1, we follow a procedure similar
to Example 2.1 and show that

Material C: K = 300 MPa, Ao = 3 mm2;
Ao

Aneck

= 1.1

41-45
330-360
180-214
190-200
190-200
80-130
350-400
70-1000
70-80
0.01-0.1
1.4-3.4
3.5-17

0.32
0.32
0.31
0.30
0.29
0.32
0.27
0.2
0.24
0.5
0.36
0.34


14

ν
0.32
0.34
0.43

G (GPa)
26-30
39-56
4.9

12

Recall that toughness is the area under the
stress-strain curve, hence the toughness for this


.

material would be given by

Toughness

approximate c = 3.3 for cold-worked copper.
However, since the Brinell hardness is in units of
kg/mm2, from Eq. (2.29) we can write
Tsteel

=

that (a) highly cold-worked metals such as these
have a much higher yield stress than the
annealed materials described in the text, and (b)
arbitrary property values are given in the
statement of the problem.

2.57 Calculate the work done in frictionless
compression of a solid cylinder 40 mm high and
15 mm in diameter to a reduction in height of
75% for the following materials: (1) 1100-O
aluminum, (2) annealed copper, (3) annealed
The fracture plane is oriented at an angle of 45◦ ,
304 stainless steel, and (4) 70-30 brass,
corresponding to a rotation of 90◦ on the Mohr’s
annealed.
circle. This corresponds to a stress state on the
The work done is calculated from Eq. (2.62) on
fracture plane of σ = −19,100 psi and τ = 19,100
p. 71 where the specific energy, u, is obtained
psi.
from Eq. (2.60). Since the reduction in height is
2.56 What is the modulus of resilience of a highly cold75%, the final height is 10 mm and the absolute
worked piece of steel with a hardness of 300
value of the true strain is
HB? Of a piece of highly cold-worked copper
with a hardness of 150 HB?
2 =90°

Referring to Fig. 2.24 on p. 55, the value of c in
Eq. (2.29) on p. 54 is approximately 3.2 for

Eq. (2.60). For example, for 1100-O aluminum,
where K is 180 MPa and n is 0.20, u is calculated 2.59 A tensile-test specimen is made of a material
as
represented by the equation
.
(a) Determine the true strain at which
= 222 MN/m3
neckingwill begin. (b) Show that it is
possible for an engineering material to
The volume is calculated as V = πr2l =
exhibit this behavior. (a) In Section 2.2.4 on
π(0.0075)2(0.04) = 7.069×10−6 m3. The work
p. 38 we noted that instability, hence
done is the product of the specific work, u, and
necking, requires the following condition to
the volume, V . Therefore, the results can be
be fulfilled:
tabulated as follows.
u

Work

Material
(MN/m3)
(Nm)
1100-O Al
222
1562
Cu, annealed
338

17, we have, as
same percent reduction, say 50%. Prove that the
the true ultimate tensile strength:
final diameters will be the same.
UTStrue = (100,000)(0.17)0.17 = 74,000 psi.
The cross-sectional area at the onset of necking
is obtained from
.

Let’s identify the shorter cylindrical specimen
with the subscript s and the taller one as t, and
their original diameter as D. Subscripts f and o
indicate final and original, respectively. Because
both specimens undergo the same percent
reduction in height, we can write

Consequently,
and from volume constancy,

Aneck = Aoe−0.17 and
the maximum load, P, is

14


.

Since
can now write


a maximum at some strain. This is seen in the plot of
σb shown below.
50,000

a
40,000

F

2

1

c

30,000

c
x

20,000

b

10,000

From the equilibrium of vertical forces and to keep the
bar horizontal, we note that 2Fa = Fb. Hence, in terms
of true stresses and instantaneous areas, we have
2σaAa = σbAb

2.65 Show that you can take a bent bar made of an
value of stress.
elastic, perfectly plastic material and straighten
• In compression testing of brittle materials,
it by stretching it into the plastic range. (Hint:
such as ceramics, when the specimen
Observe the events shown in Fig. 2.32.)
begins to fracture.
The series of events that takes place in
• If the material is susceptible to thermal
straightening a bent bar by stretching it can be
softening, then it can display such behavior
visualized by starting with a stress distribution
with a sufficiently high strain rate.
as in Fig. 2.32a on p. 61, which would represent
the unbending of a bent section. As we apply
2.63 In a disk test performed on a specimen 40-mm
tension, we algebraically add a uniform tensile
in diameter and 5 m thick, the specimen
stress to this stress distribution. Note that the
fractures at a stress of 500 MPa. What was the
change in the stresses is the same as that
load on the disk at fracture?
depicted in Fig. 2.32d, namely, the tensile stress
Equation (2.20) is used to solve this problem.
increases and reaches the yield stress, Y . The
Noting that σ = 500 MPa, d = 40 mm = 0.04 m,
compressive stress is first reduced in
and t = 5 mm = 0.005 m, we can write
magnitude, then becomes tensile. Eventually,

Thus,
Y = p2(MR)E = p2(30)(30 × 106)
or Y = 42,430 psi. Using Eq. (2.32), the strain
required to relieve the residual stress is:

Since Y = 600 MPa and E = 200 GPa, we find that
the elastic limit for this material is at an elastic
strain of

Therefore,

Y
e=

=
E

16

600 MPa
= 0.003
200 GPa


.

which is much smaller than 0.05. Following the
description in Answer 2.65 above, we find that
the strain required to straighten the bar is e =
(2)(0.003) = 0.006

tension, simple compression, equal biaxial
tension, and equal biaxial compression. Thus,
acceptable answers would include (a) wire rope,
as used on a crane to lift loads; (b) spherical
pressure vessels, including balloons and gas
storage tanks, and (c) shrink fits.
2.69 A thin-walled spherical shell with a yield stress Y
is subjected to an internal pressure p. With
appropriate equations, show whether or not the
pressure required to yield this shell depends on
the particular yield criterion used.
Here we have a state of plane stress with equal
biaxial tension. The answer to Problem 2.68
leads one to immediately conclude that both the
maximum shear stress and distortion energy
criteria will give the same results. We will now
demonstrate this more rigorously. The principal
membrane stresses are given by

and
which represents a new loading with an
additional hydrostatic pressure, p. The
distortionenergy criterion for this stress state is

σ3 = 0
Using the maximum shear-stress criterion, we
find that
σ1 − 0 = Y
hence


= 0. From Eq. 2.43b on p. 68, we find that σ2 =
σ1/2. Substituting these into the distortionenergy criterion given by Eq. (2.37) on p.64,

Since all the quantities are positive (note that in
order to produce a tensile membrane stress, the
pressure is positive as well), the longitudinal
strain is finite and positive. Thus the cylinder
becomes longer when pressurized, as it can also
be deduced intuitively.
2.73 A round, thin-walled tube is subjected to tension
in the elastic range. Show that both the thickness
and the diameter of the tube decrease as tension
increases.

and

hence
2.71 What would be the answer to Problem 2.70 if the
maximum-shear-stress criterion were used?
Because σ2 is an intermediate stress and using
Eq. (2.36), the answer would be
σ1 − 0 = Y
hence the yield stress in plane strain will be
equal to the uniaxial yield stress, Y .

2.72 A closed-end, thin-walled cylinder of original
length l, thickness t, and internal radius r is
subjected to an internal pressure p. Using the
generalized Hooke’s law equations, show the
change, if any, that occurs in the length of this

a rectangle, with its long axis in the longitudinal
direction, or (4) an ellipse? Perform this
experiment and, based on your observations,
explain the results, using appropriate equations.
Assume that the material the balloon is made of
is perfectly elastic and isotropic, and that this
situation represents a thin-walled closed-end
cylinder under internal pressure.


.

This is a simple graphic way of illustrating the
generalized Hooke’s law equations. A balloon is
a readily available and economical method of
demonstrating these stress states. It is also
encouraged to assign the students the task of
predicting the shape numerically; an example of
a valuable experiment involves partially
inflating the balloon, drawing the square, then
expanding it further and having the students
predict the dimensions of the square.
Although not as readily available, a rubber tube
can be used to demonstrate the effects of torsion
in a similar manner.

where p is the hydrostatic pressure. Thus, from
Table 2.1 on p. 32 we take values for steel of ν =
0.3 and E = 200 GPa, so that


strain is found to be

From the definition of true strain given by
Eq. (2.9) on p. 35, ln

, etc., so that
.

2.78 (a) Calculate the work done in expanding a 2mmthick spherical shell from a diameter of 100 mm
to 140 mm, where the shell is made of a material
for which
MPa. (b) Does your
answer depend on the particular yield criterion
From Eq. (2.46) on p. 68 and noting that, for this
used? Explain.
case, all three strains are equal and all three
stresses are equal in magnitude,
For this case, the membrane stresses are given
by

2.76 What is the diameter of an originally 30mmdiameter solid steel ball when it is subjected
to a hydrostatic pressure of 5 GPa?

19


.

and the strains are


rf

2.79 A cylindrical slug that has a diameter of 1 in. and
is 1 in. high is placed at the center of a 2-in.Using the pressure-volume method of work, we
diameter cavity in a rigid die. (See the
begin with the formula
accompanying figure.) The slug is surrounded
ZW=
by a compressible matrix, the pressure of which
pdV
is given by the relation
where V is
2

the volume
of the
sphere. We
integrate
this
equation
between the
limits Vo
and Vf,
noting that

psi
om

where m denotes the matrix and Vom is the
original volume of the compressible matrix.

pressure increases, thus subjecting the slug to
transverse compressive stresses on its The required compressive stress on the slug is
circumference. Hence the slug will be subjected
to
triaxial compressive stresses, with σ2 = σ3. Using We may now write the total force on the piston as
the maximum shear-stress criterion for
simplicity, we have
lb.
σ1 = σ + σ2
where σ1 is the required compressive stress on The following data gives some numerical results:
the slug, σ is the flow stress of the slug material
corresponding to a given strain, and given as
, and σ2 is the compressive stress
due to matrix pressure. Lets now determine the
matrix pressure in terms of d.
The volume of the slug is equal to π/4 and the
volume of the cavity when d = 0 is π. Hence the
original volume of the matrix is
. The
volume of the matrix at any value of d is then
,

And the following plot shows the desired results.

from which we obtain

2.80 A specimen in the shape of a cube 20 mm on each
(a) For a perfectly-elastic material as shown in
side is being compressed without friction in a
Fig 2.7a on p. 40, this expression becomes

in Fig. 2.7b, this is

We note that the volume of the specimen is constant and can be expressed as
(c) For an elastic, perfectly plastic material, this
is identical to an elastic material for
, and for
it is
where x is the lateral dimensions assuming the specimen
expands uniformly during compression. Since h = 3 mm, we
have x = 51.6 mm. Thus, the specimen touches the walls and
hence this becomes a plane-strain problem (see Fig. 2.35d
on p. 67). The absolute value of the true strain is
We can now determine the flow stress, Yf, of the
(20)(20)(20) = (h)(x)(x)

material at this strain as
(d) For a rigid, linearly strain hardening
material, the specific energy is
Yf = 70 + 30(1.90) = 127 MPa
The cross-sectional area on which the force is acting is

(e) For an elastic, linear strain hardening
material, the specific energy is identical to
Area = (20)(20)(20)/3 = 2667 mm2
an elastic material for
and for
it is
According to the maximum shear-stress criterion, we have σ1
= Yf, and thus
Force = (127)(2667) = 338 kN

strip has increased in width to 52 mm. What is
the strain in the rolling direction?
The thickness strain is

If Y = 70 MPa and σ1, σ2 = σ1/3 and σ3 =
−σ1/2 is the stress state, then

5 mm
t

=

−0.693

lo

1 mm
Thus, σ1 = 60.0 MPa. Therefore, the stress level
to initiate yielding actually increases when σ2 is
increased.
2.83 A steel plate has the dimensions 100 mm × 100
mm × 5 mm thick. It is subjected to biaxial
tension of σ1 = σ2, with the stress in the thickness
direction of σ3 = 0. What is the largest possible
change in volume at yielding, using the von
Mises criterion? What would this change in
volume be if the plate were made of copper?
From Table 2.1 on p. 32, it is noted that for steel
we can use E = 200 GPa and ν = 0.30. For a stress
state of σ1 = σ2 and σ3 = 0, the

uniaxial tension. If this material is subjected to
the stresses σ1 = 25 MPa, σ2 = 15 MPa and σ3 =
−26 MPa, will it yield? Explain.
According to the maximum shear-stress
criterion, the effective stress is given by Eq.
(2.51) on p. 69 as:
σ¯ = σ1 − σ3 = 25 − (−26) = 51 MPa
However, according to the distortion-energy
criterion, the effective stress is given by Eq.
(2.52) on p. 69 as:

or
[(350 MPa) + (350 MPa]
Since the original volume is (100)(100)(5) =
50,000 mm3, the stressed volume is 50,070 mm3,
or the volume change is 70 mm3.
For copper, we have E = 125 GPa and ν = 0.34.
Following the same derivation, the dilatation for
copper is 0.0006144; the stressed volume is
23

r(25 − 15)2 + (15 + 26)2 + (−26 − 25)2
σ¯ =
2
or ¯σ = 46.8 MPa. Therefore, the effective stress
is higher than the yield stress for the maximum
shear-stress criterion, and lower than the yield


.

70

The expression for heat is given by
Heat

=

cpρV ∆T

h1 70
Note that if the operation were
conducted in one step, the following

=

would result:

(0.3)(0.1)(0.785)(300)(778) =
5500ft-lb = 66,000 in-lb.

=
=

As was shown in Problem 2.46, this indicates
that the true strains are additive while the
engineering strains are not.

n +1

K


Therefore,
have

60. Using absolute values, we

5 mm
Based on these diameters the cross-sectional
area at the steps is calculated as:
= 7181 mm2

Solving for hf gives hf = 0.074 in.
24


.

566 mm2
As calculated in Problem 2.87,
357 and
916. Note that for 1100-O aluminum,
K = 180 MPa and n = 0.20 (see Table 2.3 on p.
37) so that Eq. (2.11) on p. 35 yields σ1 =
180(0.357)0.20 = 146.5 MPa σ2 = 180(0.916)0.20 =

Consequently,
Aneck = Aoe−0.22 and
the maximum load is
P = σA = σultAneck.


tensile strength for this material.
This solution follows the same approach as in
Example 2.1. From Eq. (2.11) on p. 35, and
recognizing that n = 0.22 and σ = 20,000 psi for
20,
or K = 28,500 psi. Therefore, the stress-strain
relationship for this material is
psi
To determine the ultimate tensile strength for
the material, realize that the strain at necking is
equal to the strain hardening exponent, or
. Therefore, σult = K(n)n = 28,500(0.22)0.22 =
20,400 psi
The cross-sectional area at the onset of necking
is obtained from

25

Since the area of each face is 400 mm2, the
stresses in the x- and y- directions are
100 MPa
200 MPa
where the negative sign indicates that the
stresses are compressive. If the Tresca criterion
is used, then Eq. (2.36) on p. 64 gives
σmax − σmin = Y = 2k = 280 MPa
It is stated that σ3 is compressive, and is
therefore negative. Note that if σ3 is zero, then
the material does not yield because σmax − σmin =
0 − (−200) = 200 MPa < 280 MPa. Therefore, σ3